Question

If $k>1$ and the determinant of the matrix $A^2$ is $k^2$, then $\left|\alpha \right|=$

$A=\left[\begin{array}{ccc}k& k\alpha & \alpha \\ 0& \alpha & k\alpha \\ 0& 0& k\end{array}\right]$

• A. $\frac{1}{k^2}$
• B. $k$
• C. $k^2$
• D. $\frac{1}{k}$

Answer 1

The correct option is D

$\frac{1}{k}$

Explanation for the correct option:

Find the value of $\left|\alpha \right|$:

The matrix $A=\left[\begin{array}{ccc}k& k\alpha & \alpha \\ 0& \alpha & k\alpha \\ 0& 0& k\end{array}\right]$, so

$\begin{array}{rcl}\left|A\right|& =& k\left(\alpha k-0\right)-ka\left(0-0\right)+a\left(0-0\right)\\ & =& \alpha k^2\end{array}$

Now,

$\begin{array}{rcl}\left|A^2\right|& =& {\alpha }^2{k}^4\\ \Rightarrow k^2& =& {\alpha }^2{k}^4\\ & \Rightarrow & \frac{1}{k^2}={\alpha }^2\\ \Rightarrow \alpha & =& \sqrt{\frac{1}{k^2}}\\ & =& \frac{1}{k}\end{array}$

Hence, option D is correct.